# Does there exist another form of the derivative for polynomials?

Let $$F : \mathbb{R}[X] \rightarrow \mathbb R[X]$$ be a linear map and let $$H \in \mathbb{R}[u,x,y,z]$$ be a polynomial. Suppose that

$$F(P \cdot Q) = H(F(P),F(Q),P,Q)$$

for all $$P, Q \in \mathbb{R}[X]$$. Two obvious solutions for $$F$$ and $$H$$ are

• $$F = I$$, the identity function, and $$H(u,x,y,z) = \alpha u x + \beta u z + \gamma yx + \delta yz$$ where $$\alpha,\beta,\gamma,\delta \in \mathbb{R}$$ and $$\alpha +\beta+\gamma+\delta = 1$$;

• $$F = D$$, the derivative, and $$H(u,x,y,z) = u z + x y$$, from the Leibniz rule.

Do there exist solutions $$F$$ and $$H$$ in which $$F$$ is not a linear combination of $$I$$ and $$D$$?

• Am I the only one who did not understand the question? – Praphulla Koushik Jun 15 at 14:06
• Is H supposed to be a polynomial? – Daniil Rudenko Jun 15 at 15:00
• It might be clearer to use $\mathbb{R}[X]$ to avoid $x$ as an indeterminate in two different rings (assuming that $H$ is indeed a polynomial). – Mark Wildon Jun 15 at 17:26
• @Dattier One can make further examples by taking $F = \lambda I + \mu D$, where $F$ is the identity and $D$ is the derivative, and $H(u,x,y,z) = \lambda (\alpha u x + \beta u z + \gamma xy + \delta y z) + \mu (u z + x y)$. I think your question is interesting, but probably it won't get the attention it deserves unless you edit it to make it clearer. In particular, you might define a `non-trivial' example to be one not of the form in this comment. – Mark Wildon Jun 16 at 9:49
• @Dattier: I have rewritten the question in the usual notation, adding in the examples from my comment. You can revert it you like, but obviously I prefer my version. – Mark Wildon Jun 16 at 11:45

It's not hard to see that $$H$$ must be of the form $$\alpha ux + \beta uz + \gamma yx + \delta yz$$ by $$\mathbb{R}$$-linearity of $$F$$ (see Jan-Cristoph Schlage-Puchta's answer for a fuller explanation).

By using symmetry of multiplication, we can assume without loss of generality that $$\beta = \gamma$$.

We have the equation $$F(P \times Q) = H(F(P), F(Q), P, Q) = \alpha F(P) F(Q) + \beta F(P) Q + \gamma P F(Q) + \delta PQ$$

Setting $$P = Q = 1$$, we get $$F(1) = \alpha F(1)^2 + \beta F(1) + \gamma F(1) + \delta$$. Degree-checking tells us that if $$F(1)$$ isn't constant, then $$\alpha = 0$$. More careful checking tells us that $$\beta + \gamma = 1, \delta = 0$$. This leads to the solution $$F_{M, R}(P) = P \times R$$, where $$R = F(1)$$. That answer aside, we assume $$F(1)$$ is constant.

I'm going to take a detour for a moment, to talk about an action of the two-dimensional nonabelian Lie group on the space of pairs $$(F, H)$$ that satisfy $$F(P \times Q) = H(F(P), F(Q), P, Q)$$. This action comes from the map $$g_{(a, b)} F = aF + b \operatorname{Id}$$. We want the equation to remain the same, which leads us to:

$$(gH)(u, x, y, z) = aH(\frac{u - by}{a}, \frac{x - bz}{a}, y, z) + b yz$$

$$= \frac{\alpha}{a} ux + (\beta - \frac{b}{a} \alpha) uz + (\gamma - \frac{b}{a} \alpha) yx + (a \delta - b \beta - b \gamma + \frac{b^2}{a} \alpha + b) yz$$

Back from the detour: we now set just $$Q = 1$$, getting $$F(P) = \alpha F(P) F(1) + \beta F(P) + \gamma P F(1) + \delta P$$. Rearranging, we get $$(1 - \alpha F(1) - \beta) F(P) = (\gamma F(1) + \delta) P$$. The obvious solutions have $$F(P) = \lambda P$$ for some $$\lambda \in \mathbb{R}$$, which have been discussed in other answers; otherwise, we must have $$1 - \alpha F(1) - \beta = \gamma F(1) + \delta = 0$$.

We now split into two cases: either $$\alpha = 0$$, or $$\alpha \neq 0$$. If $$\alpha = 0$$, then $$\beta = \gamma = 1$$. Using the action of the group, we can reduce to the case that $$\delta = 0$$, implying that $$F(P \times Q) = F(P) Q + P F(Q)$$, i.e. that $$F$$ is a derivation. This can only happen when $$F(P) = R \times \partial P$$ for some polynomial $$R$$. Undoing the group action, we get $$F_{D, R, \lambda}(P) = \lambda P + R \times \partial P$$.

If $$\alpha \neq 0$$, then we can use the action of the group to reduce to the case $$\alpha = 1, \beta = \gamma = 0$$. But then $$\delta = 0$$, giving us the equation $$F(P \times Q) = F(P) F(Q)$$ - in other words, $$F$$ is a homomorphism. Homomorphisms from $$\mathbb{R}[X]$$ are compositions: $$F(P) = P \circ R$$ for some polynomial $$R$$. Undoing the group action, we get $$F_{H, R, \lambda} = \lambda P + P \circ R$$.

So all solutions $$(F, H)$$ are of the forms:

1) $$F(P) = \lambda P, \frac{1}{\lambda} \alpha + \beta + \gamma + \lambda \delta = 1$$

2) $$F(P) = \lambda P + R \times \partial P, \alpha = 0, \beta = \gamma = 1, \delta = \lambda$$

3) $$F(P) = \lambda P + c (P \circ R)$$ with coefficients that aren't difficult to determine using the group action.

4) $$F(P) = P \times R, \alpha = 0, \beta = \gamma = \frac{1}{2}, \delta = 0$$

If I understand everything correctly, another simple solution is $$F(P) = P(a)$$ (evaluation in a given $$a\in\mathbb{R}$$), with $$H(u,x,y,z)=ux$$.

• I think any substitution works: for any polynomial $R$, $F_R(P) = P \circ R, H = ux$. – user44191 Jun 16 at 14:00
• @user44191 Good point. And also evaluating the derivative works. – Federico Poloni Jun 16 at 14:44

As there are several possibilities for $$F$$, here is an attempt at determining $$H$$.

Using the linearity of $$F$$ we have $$H(\lambda x, y, \lambda u, v)=\lambda H(x, y, u, v)$$. Taking the derivative with respect to $$\lambda$$ we obtain $$xH_x+uH_u=H$$. The map $$x\partial x+u\partial u$$ is linear and maps a monomial $$x^ay^bu^cv^d$$ to $$(a+c)x^ay^bu^cv^d$$, hence, every monomial with a non-zero coefficient satisfies $$a+c=1$$.

The same argument applies to $$y$$ and $$v$$, and we conclude that every possible polynomial $$H$$ is of the form $$\alpha xy+\beta xv+\gamma yu + \delta uv$$. On the other hand we have that if $$F=\lambda\mathrm{id}$$, then $$H(x,y,u,v)=\alpha xy+\beta xv+\gamma yu + \delta uv$$ with $$\lambda\alpha+\beta+\gamma+\frac{\delta}{\lambda}=1$$ is a possible polynomial $$H$$. In particular, all tuples $$(\alpha, \beta, \gamma, \delta)$$ with $$\alpha\delta<0$$ are possible. So further restrictions on the possible shape of $$H$$ are only minor.

• I think I'm missing something; how do you get that all tuples following the inequality are possible? It seems to me all you proved is that tuples following the equality are possible (for $F = \lambda Id$). – user44191 Jun 16 at 17:50
• @user44191: If $\alpha>0$ and $\delta<0$, then $\lambda\mapsto \lambda\alpha+\frac{\delta}{\lambda}$ is a continuous function tending to $\infty$ for $\lambda\rightarrow\infty$ and to $-\infty$ for $\lambda\rightarrow 0$. So for any given $\beta, \gamma$, there is some $\lambda$, such that $\lambda\alpha+\beta+\gamma+\frac{\delta}{\lambda}=1$. So for any such tuple there is some function $F$ (namely $\lambda\mathrm{id}$ for a suitable $\lambda$), for which this particular $H$ satisfies the functional equation. – Jan-Christoph Schlage-Puchta Jun 27 at 15:52